如何在Java中实现二叉搜索树（ binary search tree）？

<b>import</b> java.util.Stack;

<font><i>/**
 * Java Program to implement a binary search tree. A binary search tree is a
 * sorted binary tree, where value of a node is greater than or equal to its
 * left the child and less than or equal to its right child.
 * 
 * @author WINDOWS 8
 *
 */</i></font><font>
<b>public</b> <b>class</b> BST {

    <b>private</b> <b>static</b> <b>class</b> Node {
        <b>private</b> <b>int</b> data;
        <b>private</b> Node left, right;

        <b>public</b> Node(<b>int</b> value) {
            data = value;
            left = right = <b>null</b>;
        }
    }

    <b>private</b> Node root;

    <b>public</b> BST() {
        root = <b>null</b>;
    }

    <b>public</b> Node getRoot() {
        <b>return</b> root;
    }

    </font><font><i>/**
     * Java function to check if binary tree is empty or not
     * Time Complexity of this solution is constant O(1) for
     * best, average and worst case. 
     * 
     * @return true if binary search tree is empty
     */</i></font><font>
    <b>public</b> <b>boolean</b> isEmpty() {
        <b>return</b> <b>null</b> == root;
    }

    
    </font><font><i>/**
     * Java function to return number of nodes in this binary search tree.
     * Time complexity of this method is O(n)
     * @return size of this binary search tree
     */</i></font><font>
    <b>public</b> <b>int</b> size() {
        Node current = root;
        <b>int</b> size = 0;
        Stack<Node> stack = <b>new</b> Stack<Node>();
        <b>while</b> (!stack.isEmpty() || current != <b>null</b>) {
            <b>if</b> (current != <b>null</b>) {
                stack.push(current);
                current = current.left;
            } <b>else</b> {
                size++;
                current = stack.pop();
                current = current.right;
            }
        }
        <b>return</b> size;
    }


    </font><font><i>/**
     * Java function to clear the binary search tree.
     * Time complexity of this method is O(1)
     */</i></font><font>
    <b>public</b> <b>void</b> clear() {
        root = <b>null</b>;
    }

}
</font>